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- čas přidán 21. 09. 2023
- #GFG #POTD #geeksforgeeks #problemoftheday
In this video, I will be discussing the Equilibrium Point Problem.
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Given an array A of n positive numbers. The task is to find the first equilibrium point in an array. Equilibrium point in an array is a position such that the sum of elements before it is equal to the sum of elements after it.
Note: Return equilibrium point in 1-based indexing. Return -1 if no such point exists.
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Very good explaination
Thanks Riya. Glad that you liked it.
At 8:25,
Is the rightSum = totalSum - a[i] - leftSum ?
Yes please go through full video u will get it
@@PlacementsReady but it wont work it we have first position as equilibrium point for e.g: 1,0 so we need to use rightsum = totalsum - a[i], leftsum += a[i] and consdition to return is if leftsum == rightsum-a[i]
Two pointer method is more optimised, only takes O(n) time instead of O(2n)
can you explain little bit...about two pointer method
@farhanmuhamed392 how are you solving this question using two pointer approach?
@@PlacementsReady
int equilibriumPoint(long long a[], int n) {
int left=0, right=n-1;
long long sumLeft=a[left], sumRight=a[right];
while(left != right){
if(sumLeft > sumRight){
right--;
sumRight = sumRight + a[right];
}
else{
left++;
sumLeft = sumLeft + a[left];
}
}
if( sumLeft == sumRight) return left+1;
else return -1;
}
@@cosmosXverse Declare 4 variables left, right, leftSum, rightSum
left and right are pointers , and leftSum and rightSum keep track sum of value at current pointers
int equilibriumPoint(long long a[], int n) {
int left=0, right=n-1;
long long sumLeft=a[left], sumRight=a[right];
while(left != right){
if(sumLeft > sumRight){
right--;
sumRight = sumRight + a[right];
}
else{
left++;
sumLeft = sumLeft + a[left];
}
}
if( sumLeft == sumRight) return left+1;
else return -1;
}
@@farhanmuhamed392 bro it is failing for the test case 1 2 0 3 it expected output is 3 but it is printing -1